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CIE 9709 2025 November Paper 52 Q3

A Level / CIE / S1

CIE 9709 2025 Nov Paper 52 Paper · Question 3

题目

Problem

The heights of the 124 Senior members of the Giraffes basketball club are normally distributed with mean 187.4 cm187.4\,\mathrm{cm} and standard deviation 6.4 cm6.4\,\mathrm{cm}.

(a) How many members of the club would you expect to have heights within 5 cm5\,\mathrm{cm} of the mean?

(4)

The heights of the Junior members of the Giraffes club are normally distributed with mean 172.7 cm172.7\,\mathrm{cm} and standard deviation σ cm\sigma\,\mathrm{cm}. 23% of these members have height less than 170.3 cm170.3\,\mathrm{cm}.

(b) Find the value of σ\sigma.

(3)
题目中文翻译

Giraffes 篮球俱乐部 124 名 Senior 组成员的身高服从正态分布,平均数为 187.4 cm187.4\,\mathrm{cm},标准差为 6.4 cm6.4\,\mathrm{cm}。

(a) 预计该俱乐部有多少名成员的身高与平均数相差不超过 5 cm5\,\mathrm{cm}?

(4)

Giraffes 俱乐部 Junior 组成员的身高服从正态分布,平均数为 172.7 cm172.7\,\mathrm{cm},标准差为 σ cm\sigma\,\mathrm{cm}。其中 23% 的成员身高低于 170.3 cm170.3\,\mathrm{cm}。

(b) 求 σ\sigma 的值。

(3)
原卷通用作答要求

Answer all questions. You should use a calculator where appropriate.

You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程;仅给出计算器答案而没有过程,不予评分。除非题目另有精度要求,非精确数值答案保留 3 位有效数字,以度为单位的角度保留 1 位小数。

解答

(a)

解法一

思路

展开

离均值5厘米内的区间两端对称。标准化后求中央概率,再乘124给出预计人数。

答题过程

展开

Let H∼N(187.4,6.42)H\sim\mathrm{N}(187.4,6.4^2) denote a Senior member’s height in centimetres. Then

P(∣H−187.4∣<5)= P(−56.4<Z<56.4)= 2Φ(0.78125)−1= 0.565344….\begin{align*} \mathrm{P}(|H-187.4|<5) =&\,\mathrm{P}\biggl(-\frac5{6.4}<Z<\frac5{6.4}\biggr)\\[4mm] =&\,2\Phi(0.78125)-1\\[4mm] =&\,0.565344\ldots. \end{align*}

The expected number is 124(0.565344…)=70.1027…124(0.565344\ldots)=70.1027\ldots, giving 70 members.

(b)

解法一

思路

展开

170.3低于均值,对应左尾概率0.23的负标准分数。标准化后求正的标准差。

答题过程

展开

For a Junior member,

170.3−172.7σ=Φ−1(0.23)=−0.738846….\frac{170.3-172.7}{\sigma}=\Phi^{-1}(0.23) =-0.738846\ldots.

Hence

σ=−2.4−0.738846…=3.248305…≈3.25.\sigma=\frac{-2.4}{-0.738846\ldots} =3.248305\ldots \approx\boxed{3.25}.